N-Queens

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
思路：八皇后问题一直是一个经典的回溯算法，递归算法解决。这道题受字符串的全排列的启发——定义一个数组data[n],数组中第i个数组表示位于第i行的皇后的列号。先把数组data的n个数字初始化分别用0~n-1初始化，接下来就是对数组data做全排列。我们只需要判断每一个排列对应的n个皇后是不是在同一对角线上，也就是对于数组的两个下标i和j，是不是j-i=abs(data[j]-data[i]).

class Solution {
public:
    bool check(vector<int> &data,int n)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                if((j-i)==abs(data[j]-data[i]))
                {
                    return false;
                }
            }
        }
        return true;
    }
    void Permutation(vector<vector<string> > &result,vector<int> &data,int pBegin,int n)
    {
        if(pBegin==n)
        {
            if(true==check(data,n))
            {
                vector<string> pTemp;
                for(int i=0;i<n;i++)
                {
                    string row(n,'.');
                    row[data[i]]='Q';
                    pTemp.push_back(row);
                }
                result.push_back(pTemp);
            }
        }
        else
        {
            for(int i=pBegin;i<n;i++)
            {
                swap(data[pBegin],data[i]);
                Permutation(result,data,pBegin+1,n);
                swap(data[pBegin],data[i]);
            }
        }
    }
    void Permutation(vector<vector<string> > &result,int n)
    {
        vector<int> data(n);//
        for(int i=0;i<n;i++)
            data[i]=i;
        Permutation(result,data,0,n);
    }
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string> > result;
        result.clear();
        Permutation(result,n);
        return result;
    }
};